Answer to Question #320861 in Statistics and Probability for Michelle

Question

Box A box B both contain the numbers 1,2,3 and 4 contrast the probability mass function and draw the histogram of the sum when one number from each box is take at a time with replacement

Answer 1

$X_1-number\,\,from\,\,the\,\,1st\,\,box\\X_2-number\,\,from\,\,the\,\,2nd\,\,box\\S=X_1+X_2\in \left\{ 2,3,4,5,6,7,8 \right\} \\P\left( S=2 \right) =P\left( X_1=1,X_2=1 \right) =\frac{1}{4}\cdot \frac{1}{4}=0.0625\\P\left( S=3 \right) =P\left( X_1=1,X_2=2 \right) +P\left( X_1=2,X_2=1 \right) =2\cdot \frac{1}{4}\cdot \frac{1}{4}=0.125\\P\left( S=4 \right) =P\left( X_1=1,X_2=3 \right) +P\left( X_1=2,X_2=2 \right) +P\left( X_1=3,X_2=1 \right) =3\cdot \frac{1}{4}\cdot \frac{1}{4}=0.1875\\P\left( S=5 \right) =P\left( X_1=1,X_2=4 \right) +P\left( X_1=2,X_2=3 \right) +P\left( X_1=3,X_2=2 \right) +P\left( X_1=4,X_2=1 \right) =4\cdot \frac{1}{4}\cdot \frac{1}{4}=0.25\\P\left( S=6 \right) =P\left( X_1=2,X_2=4 \right) +P\left( X_1=3,X_2=3 \right) +P\left( X_1=4,X_2=2 \right) =3\cdot \frac{1}{4}\cdot \frac{1}{4}=0.1875\\P\left( S=7 \right) =P\left( X_1=3,X_2=4 \right) +P\left( X_1=4,X_2=3 \right) =2\cdot \frac{1}{4}\cdot \frac{1}{4}=0.125\\P\left( S=8 \right) =P\left( X_1=4,X_2=4 \right) =\frac{1}{4}\cdot \frac{1}{4}=0.0625$